Let vec{a}=3 hat{i}+hat{j}-2 hat{k},vec{b}=-5 | vedclass

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(D) Given vectors are $\vec{a}=3 \hat{i}+\hat{j}-2 \hat{k}$,$\vec{b}=-5 \hat{i}+7 \hat{j}$,and $\vec{c}=3 \hat{i}+y \hat{j}$.First,calculate $\vec{a}-

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(D) Given vectors are $\vec{a}=3 \hat{i}+\hat{j}-2 \hat{k}$,$\vec{b}=-5 \hat{i}+7 \hat{j}$,and $\vec{c}=3 \hat{i}+y \hat{j}$.First,calculate $\vec{a}-\vec{b}+\vec{c}$:$\vec{a}-\vec{b}+\vec{c} = (3 \hat{i}+\hat{j}-2 \hat{k}) - (-5 \hat{i}+7 \hat{j}) + (3 \hat{i}+y \hat{j})$$= (3+5+3) \hat{i} + (1-7+y) \hat{j} - 2 \hat{k}$$= 11 \hat{i} + (y-6) \hat{j} - 2 \hat{k}$.Now,find the magnitude $|\vec{a}-\vec{b}+\vec{c}| = \sqrt{11^2 + (y-6)^2 + (-2)^2} = \sqrt{121 + (y-6)^2 + 4} = \sqrt{125 + (y-6)^2}$.Given that $|\vec{a}-\vec{b}+\vec{c}| = \sqrt{141}$,we have:$\sqrt{125 + (y-6)^2} = \sqrt{141}$$125 + (y-6)^2 = 141$$(y-6)^2 = 141 - 125 = 16$$y-6 = \pm 4$.Thus,$y_1 = 6+4 = 10$ and $y_2 = 6-4 = 2$.The value of $|y_1-y_2| = |10-2| = 8$.

Let $\vec{a}=3 \hat{i}+\hat{j}-2 \hat{k}$,$\vec{b}=-5 \hat{i}+7 \hat{j}$,and $\vec{c}=3 \hat{i}+y \hat{j}$ be three vectors such that $|\vec{a}-\vec{b}+\vec{c}|=\sqrt{141}$. If $y_1$ and $y_2$ are the values of $y$ satisfying the given condition,then $|y_1-y_2|=$

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